Question

6. In doing the neutralization (using the same calorimeter as in problem 5) of HF with NaOH; 50.0 mL of 2.08 M NaOH at 22.2°C are mixed with 50.0 ml of 2.08 M HF at 22.2°C. After the two solutions were mixed, allow to react, and come to equilibrium temperature; the final temperature was found to be 33.2°C. The resulting solution weighed 100.791 g. In order to obtain the specific heat of the resulting NaF solution: 50 mL (51.264 g) of distilled water at 64.3°C were added to 52.177 g of the NaF solution from the resulting solution at 22.2°C in the same calorimeter. After equilibrium had been reached, the final tempera- ture was found to be 39.7°C. Calculate (a) the specific heat (cal/g - °C) of the NaF solution. (b) g. (heat of neutralization for the amount used in this experi- ment) in calories, and (c) AH, (enthalpy of neutralization of HF with NaOH) in the units of (1) cal/mol, (2) kcal/mol, and (3) kJ/mol.

Answer 1

(a) Heat lost by water = heat gained by NaF

51.264 g * 1 cal/g.^oC * (64.3-39.7) ^oC = 52.177 g * C * (39.7-22.2) ^oC

i.e. The specific heat of NaF solution (C) = 1.38 cal/g.^oC

(b) q_n = 100.791 g * 1.38 cal/g.^oC * (33.2-22.2) ^oC = 1530 cal

(c) $\Delta$H_n = 1530 cal/(100.791 g/42 g/mol) = 637.6 cal/mol or 0.6376 kcal/mol or 2.67 kJ/mol